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At what distance from a long, straight wire carrying a current of 8.7 A is the magnetic field due to the wire equal to the strength of Earth’s field, approximately 5.4 × 10−5 T? The permeabilty of free space is 1.25664 × 10−6 T · m/A. Answer in units of cm.

Respuesta :

Answer:

r= 3.2 cm

Explanation:

Given that

I= 8.7 A

B= 5.4 x 10⁻⁵ T

μo=1.25664 x 10⁻⁶

We know that magnetic filed in wire at a distance r given as

[tex]B=\dfrac{\mu_oI}{2\pi r}[/tex]

[tex]r=\dfrac{\mu_oI}{2\pi B}[/tex]

By putting the values

[tex]r=\dfrac{1.25664\times 10^{-6}\times 8.7}{2\times \pi \times 5.4\times 10^{-5}}\ m[/tex]

r=0.032 m

r= 3.2 cm

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